README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1900-1999/1968.Array%20With%20Elements%20Not%20Equal%20to%20Average%20of%20Neighbors/README_EN.md	1499	Weekly Contest 254 Q2	Greedy Array Sorting

1968. Array With Elements Not Equal to Average of Neighbors

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Description

You are given a 0-indexed array nums of distinct integers. You want to rearrange the elements in the array such that every element in the rearranged array is not equal to the average of its neighbors.

More formally, the rearranged array should have the property such that for every i in the range 1 <= i < nums.length - 1, (nums[i-1] + nums[i+1]) / 2 is not equal to nums[i].

Return any rearrangement of nums that meets the requirements.

 

Example 1:

Input: nums = [1,2,3,4,5]
Output: [1,2,4,5,3]
Explanation:
When i=1, nums[i] = 2, and the average of its neighbors is (1+4) / 2 = 2.5.
When i=2, nums[i] = 4, and the average of its neighbors is (2+5) / 2 = 3.5.
When i=3, nums[i] = 5, and the average of its neighbors is (4+3) / 2 = 3.5.

Example 2:

Input: nums = [6,2,0,9,7]
Output: [9,7,6,2,0]
Explanation:
When i=1, nums[i] = 7, and the average of its neighbors is (9+6) / 2 = 7.5.
When i=2, nums[i] = 6, and the average of its neighbors is (7+2) / 2 = 4.5.
When i=3, nums[i] = 2, and the average of its neighbors is (6+0) / 2 = 3.

 

Constraints:

• 3 <= nums.length <= 105
• 0 <= nums[i] <= 105

Solutions

Solution 1: Sorting

Since the elements in the array are distinct, we can first sort the array, then divide the array into two parts. Place the first half of the elements in the even positions of the answer array, and the second half of the elements in the odd positions of the answer array. In this way, for each element, its two adjacent elements will not be equal to its average value.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.

Python3

class Solution:
    def rearrangeArray(self, nums: List[int]) -> List[int]:
        nums.sort()
        n = len(nums)
        m = (n + 1) // 2
        ans = []
        for i in range(m):
            ans.append(nums[i])
            if i + m < n:
                ans.append(nums[i + m])
        return ans

Java

class Solution {
    public int[] rearrangeArray(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        int m = (n + 1) >> 1;
        int[] ans = new int[n];
        for (int i = 0, j = 0; i < n; i += 2, j++) {
            ans[i] = nums[j];
            if (j + m < n) {
                ans[i + 1] = nums[j + m];
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> rearrangeArray(vector<int>& nums) {
        ranges::sort(nums);
        vector<int> ans;
        int n = nums.size();
        int m = (n + 1) >> 1;
        for (int i = 0; i < m; ++i) {
            ans.push_back(nums[i]);
            if (i + m < n) {
                ans.push_back(nums[i + m]);
            }
        }
        return ans;
    }
};

Go

func rearrangeArray(nums []int) (ans []int) {
	sort.Ints(nums)
	n := len(nums)
	m := (n + 1) >> 1
	for i := 0; i < m; i++ {
		ans = append(ans, nums[i])
		if i+m < n {
			ans = append(ans, nums[i+m])
		}
	}
	return
}

TypeScript

function rearrangeArray(nums: number[]): number[] {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    const m = (n + 1) >> 1;
    const ans: number[] = [];
    for (let i = 0; i < m; i++) {
        ans.push(nums[i]);
        if (i + m < n) {
            ans.push(nums[i + m]);
        }
    }
    return ans;
}